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899. Orderly Queue

  • Time: $O(n^2)$
  • Space: $O(n)$
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class Solution {
 public:
  string orderlyQueue(string s, int k) {
    if (k > 1) {
      ranges::sort(s);
      return s;
    }

    string ans = s;

    for (int i = 1; i < s.length(); ++i)
      ans = min(ans, s.substr(i) + s.substr(0, i));

    return ans;
  }
};

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class Solution {
  public String orderlyQueue(String s, int k) {
    if (k > 1) {
      char[] chars = s.toCharArray();
      Arrays.sort(chars);
      return String.valueOf(chars);
    }

    String ans = s;

    for (int i = 1; i < s.length(); ++i) {
      String t = s.substring(i) + s.substring(0, i);
      if (ans.compareTo(t) > 0)
        ans = t;
    }

    return ans;
  }
}

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class Solution:
  def orderlyQueue(self, s: str, k: int) -> str:
    return (''.join(sorted(s)) if k > 1
            else min(s[i:] + s[:i] for i in range(len(s))))